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Green's Theorem Math 32B Discussion Session Session 8 Notes August 30, 2018 Today we'll discuss our first integral theorem. Green's theorem relates the work done by a 2 vector field on the boundary of a region in R to the integral of the curl of the vector field across that region. We'll also discuss a flux version of this result. Note. As with the past few sets of notes, these contain a lot more details than we'll actually discuss in section. Green's theorem Very shortly after you learned how to integrate single-variable functions, you learned the Funda- mental Theorem of Calculus | the way most integration actually gets done. The version you probably found yourself using most frequently says that b f(b) − f(a) = f 0(t)dt (1) ˆa for sufficiently nice functions f. Our primary use for this equation is that it allows us to evaluate definite integrals. If we find an antiderivative for the function we're trying to integrate, we can evaluate this antiderivative at the endpoints of our interval and compute the difference. In theory, we could also use (1) to compute this difference. Given a function f, perhaps we want to know how much f(x) changes as x varies from a to b. That is, we want to know how much elevation the graph of f gains between a and b: Equation (1) says that this value is given by the sum of the instantaneous changes in f. Even though f is sometimes increasing and sometimes decreasing, the total change in f between a and b is given by summing these instantaneous changes. The Fundamental Theorem of Calculus has a general form that we now want to mimic. It says that the behavior of our function f on the boundary of the interval [a; b] | that is, at the endpoints a and b | is related to the behavior of the derivative of f across the interval. Let's recall how we defined the line integral of a vector field F along a curve C. We imagined a thin wire submersed in some fluid, and on this wire was a bead which the fluid would push along the wire. We defined 1 Figure 1: The curve C = @D. the line integral C F · dr to be the total work done by the fluid as it pushed the bead along the wire, and determined´ that this must equal C F · Tds, where T is a positively oriented unit tangent vector field to the curve C. ´ Let's restrict our attention now to closed curves C. In a simply connected domain, a closed curve can be thought of as the boundary of some region. That is, the closed curve is to some region as the endpoints are to an interval. If F is a conservative vector field, we know that the work done by F around C is zero: C F·dr = 0. But what about when F is not conservative? Will there always be a nice way to compute¸ C F · dr? Can we compare the behavior of F along C to the behavior of some derivative of F along¸ a region for which C is the boundary? This is a problem we'd like to address today. To perform a na¨ıve analysis, we'll restrict our attention even further to closed curves of a specific type. Let's consider a rectangle D = f(x; y): a ≤ x ≤ b; c ≤ y ≤ dg which lies in the domain of a vector field F on R2, and we take our curve C to be the boundary of this rectangle, written C = @D. A picture of our setup can be seen in Figure1 As with the line integral along C, we would like to compute the work done by F along the curve @D, oriented counterclockwise, but this time we'll take a different approach. Consider the work done along the bottom line of our rectangle | the work done from (a; c) to (b; c). At each point along this line, our positively-oriented unit tangent vector T is the vector h1; 0i, so the quantity F · T is given by F · T = hF1;F2i · h1; 0i = F1: We find the work done along the bottom line by integrating with respect to x (this is the variable that changes as we move along this line): b work done along bottom line = F1(x; c)dx: ˆa 2 Next, consider the work done along the top line of the rectangle | the line from (a; d) to (b; d). This time our positively-oriented tangent vectors are pointing to the left, so we have T = h−1; 0i, meaning that F · T = −F1. So b work done along top line = −F1(x; d)dx: ˆa This means that the total work done along the top and bottom lines is given by b work done along top and bottom lines = (F1(x; c) − F1(x; d))dx: ˆa Now because we're looking at a rectangle which lies in the domain of F, we can use the Fundamental Theorem of Calculus to rewrite the quantity inside of this integral, for each a ≤ x ≤ b: c d @F1 @F1 F1(x; c) − F1(x; d) = (x; y)dy = − (x; y)dy: ˆd @y ˆc @y Substituting this in, we see that b d @F work done along top and bottom lines = − 1 (x; y)dydx: (2) ˆa ˆc @y A similar analysis works for the left and right lines. On the right, T = h0; 1i, so F · T = F2, while on the left T = h0; −1i and F · T = −F2. Then d d b @F2 work along left and right lines = (F2(b; y) − F2(a; y))dy = (x; y)dxdy: (3) ˆc ˆc ˆa @x Of course we find the total work done along @D by summing the values in (2) and (3). Using Fubini's theorem to interchange the order of integration, we see that the total work is given by b d @F d b @F b d @F @F − 1 (x; y)dydx + 2 (x; y)dxdy = 2 − 1 dydx: ˆa ˆc @y ˆc ˆa @x ˆa ˆc @x @y That is, @F @F F · dr = 2 − 1 dA: (4) ˛@D ¨D @x @y We call the result in (4) Green's Theorem, and it can be used to simplify a lot of calculations. Notice that the integrand here is precisely the z-coordinate of curl(F), where we consider F to be a vector field in three space. So in words Green's theorem is saying that the work done along @D is given by integrating the curl of F over D. Figure2 attempts to make this seem geometrically reasonable. If F is spinning in the counterclockwise direction near @D, this will generate positive work along @D. We have a helpful reality check here: if F is a conservative vector field, then the line integral of F about the closed loop @D must be zero. But we also know that conservative vector fields have zero curl, so Green's theorem gives zero, as desired. 3 Figure 2: Counterclockwise curl leads to positive work along @D. Note. It's important here that D lies entirely within the domain of F | otherwise, we might not be able to use the FTC. This subtlety is related to the requirement that a vector field’s domain be simply connected in order for the curl test for conservative vector fields to work. If D were not simply connected, we could not always apply the FTC, and the situation would become much hairier. Note. Despite the fact that we've only given an explanation for Green's theorem in the case that D is a rectangle, the equation continues to hold as long as D is a region in the domain of F whose boundary consists of a finite number of simple, closed curves, and we orient these curves so that D is always to the left. 2 Example. (Section 18.1, Exercise 8) Compute C(ln x+y)dx−x dy, where C is the rectangle with vertices (1; 1); (3; 1); (1; 4); and (3; 4), oriented clockwise.¸ (Solution) In our symbolic notation, we're being asked to compute C F · dr, where F = hln x + y; −x2i. Since we don't like integrating terms such as ln x, this is a¸ very difficult line integral to compute a priori. Green's theorem simplifies it quite a bit though, since @F @F 2 = −2x and 1 = 1: @y @y Since C is the boundary of the rectangle D = f(x; y) : 1 ≤ x ≤ 3; 1 ≤ y ≤ 4g, we have 3 4 (ln x + y)dx − x2dy = (−2x − 1)dA = (−2x − 1)dydx ˛C ¨D ˆ1 ˆ1 3 2 3 = −3 (2x + 1)dx = −3 x + x 1 = −30: ˆ1 ♦ 4 So we see that Green's theorem is a nice tool for turning difficult line integrals into double integrals. We can also use Green's theorem to go the other direction, turning a double integral into a line integral. In particular, it's occasionally helpful to use Green's theorem to compute the area of a region by rewriting the double integral D 1dA as a line integral along @D. The trick is to find @F2 @F1 ˜ a vector field F so that @x − @y = 1. Three such vector fields are given by 1 1 F(x; y) = h0; xi; F(x; y) = h−y; 0i; F(x; y) = h− y; xi; 2 2 meaning that area can be written as 1 Area = 1dA = xdy = −ydx = xdy − ydx: ¨D ˛@D ˛@D 2 ˛@D The flux version of Green's theorem Recall that the divergence of a vector field is a measure of the extent to which the vector field is expanding or contracting at a given point.
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